General expressions for $\mathcal{L}(n)=\int_{0}^{\infty}\operatorname{Ci}(x)^n\text{d}x$

Question

Define $$\operatorname{Ci}(x)=-\int_{x}^{ \infty} \frac{\cos(y)}{y}\text{d}y.$$ It is easy to show $$ \mathcal{L}(1)=\int_{0}^{\infty}\operatorname{Ci}(x)\text{d}x=0 $$ and $$\mathcal{L}(2)=\int_{0}^{\infty}\operatorname{Ci}(x)^2\text{d}x =\frac{\pi}{2}.$$ $\mathcal{L}(3),\mathcal{L}(4)$ is a little bit non-trivial. We have two claims(take a look here to find more details): $$\begin{aligned} &\mathcal{L}(3)=-\frac{3\pi}{2}\ln2 \\ &\mathcal{L}(4)=3\pi\operatorname{Li}_2 \left ( \frac{2}{3} \right )+\frac{3\pi}{2}\ln^23 \end{aligned}$$ Where $\operatorname{Li}$ are polylogarithms, they are defined by $\displaystyle{\operatorname{Li}_n(z) =\sum_{k=1}^{\infty} \frac{z^k}{k^n}}$ for $|z|<1$.
$\mathcal{L}(5)$ is much more non-trivial. We have $$ \mathcal{L}(5)=\int_{0}^{\infty}\operatorname{Ci}(x)^5\text{d}x =-\frac{15\pi^3}{8}\ln(2)-\frac{15\pi}{2}\ln(2)^3 -\frac{45\pi}{4}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2) -\frac{45\pi}{4}\operatorname{Li}_3\left ( \frac{1}{4} \right ) -\frac{15\pi}{16}\zeta(3). $$ Where $\zeta(n)=\operatorname{Li}_n(1)$ for $\Re(n)>1$.
My question: How can we find alternate generalizations? I believe that $\mathcal{L}(6)$ can be expressed by using ordinary polylogarithms($\mathcal{L}(7)$ seems impossible). We can also find the closed-forms of following integrals: $$\int_{0}^{\infty}\operatorname{Ci}(x)^4\cos(x)\text{d}x,\int_{0}^{\infty}\operatorname{Ci}(x)^2\frac{\operatorname{Si}(2x)}{x} \cos(x)\text{d}x$$ where $\displaystyle{\operatorname{Si}(x)=\int_{0}^{x} \frac{\sin(t)}{t}\text{d}t}.$

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Update 1: Define $\operatorname{si}(x)+\operatorname{Si}(x)=\frac{\pi}{2}$. Here are some results: $$\begin{aligned} &\int_{0}^{\infty}\operatorname{si}(x)\text{d}x=1\\ &\int_{0}^{\infty}\operatorname{si}(x)^2\text{d}x=\frac{\pi}{2}\\ &\int_{0}^{\infty}\operatorname{si}(x)^3\text{d}x=\frac{\pi^2}{4} -\frac{3}{2}\ln^22-\frac{3}{4} \operatorname{Li}_2\left ( \frac{1}{4} \right )\\ &\int_{0}^{\infty}\operatorname{si}(x)^4\text{d}x= \frac{\pi^3}{4} -3\pi\ln^22-\frac{3\pi}{2} \operatorname{Li}_2\left ( \frac{1}{4} \right ) \end{aligned}$$

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Update 2: A useful fourier transform $$\int_{0}^{\infty}\operatorname{Ci}(x)^3\cos(a x)\text{d}x =\begin{cases} \color{Red}{\frac{\pi \text{Li}_2\left(\frac{1-a}{3}\right)}{4 a}}+\frac{\pi \text{Li}_2\left(\frac{a-1}{a-2}\right)}{2 a}+\frac{\pi \text{Li}_2\left(\frac{a+1}{3 (a-1)}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a-1}{a+1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+1}{a+2}\right)}{2 a}-\frac{\pi \text{Li}_2\left(\frac{a+1}{3}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+1}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a-1}{3 (a+1)}\right)}{4 a}+\frac{\pi \log ^2(2-a)}{4 a}-\frac{\pi \log ^2(a+2)}{4 a}+\frac{\pi \log (3) \log (a-1)}{4 a}+\frac{\pi \log (3) \log \left(\frac{a+2}{a+1}\right)}{4 a}-\frac{\pi \log (3) \log (a-2)}{4 a}-\frac{\pi \log (3) \tanh ^{-1}\left(\frac{a}{2}\right)}{2 a} & (0\le a\le1),\\ \color{Red}{\frac{\pi \text{Li}_2\left(\frac{a^2}{a^2-1}\right)}{4 a}}+\frac{\pi \log \left(-\frac{a}{a+1}\right) \log \left(\frac{1}{1-a^2}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{a}{2}\right)}{2 a}+\frac{\pi \text{Li}_2(1-a)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (1-a)}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{3}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{1}{a}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (a+1)}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}-\frac{\pi \text{Li}_2\left(-\frac{1}{2}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{1}{1-a}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(-\frac{1}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2(-a)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{1}{a+1}\right)}{4 a}-\frac{3 \pi \text{Li}_2\left(\frac{a}{a+1}\right)}{4 a}-\frac{7 \pi ^3}{24 a}+\frac{3 \pi \log ^2(2)}{8 a}+\frac{\pi \log ^2(a)}{8 a}-\frac{\pi \log ^2(a+1)}{2 a}+\frac{\pi \log (2) \log (a-1)}{4 a}+\frac{\pi \log (2) \log (a+1)}{4 a}-\frac{\pi \log (2) \log (a)}{2 a}-\frac{\pi \log (2) \log (a+2)}{2 a}+\frac{\pi \log \left(\frac{a+2}{a+1}\right) \log \left(\frac{1}{(a+1)^2}\right)}{4 a}+\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}+\frac{\pi \log (3) \log (a+2)}{4 a}+\frac{\pi \log (a) \log (a+2)}{2 a}-\frac{\pi \log (a) \log (a+1)}{2 a}-\frac{i \pi ^2 \log \left(\frac{1}{1-a}\right)}{4 a}-\frac{\pi \log (3) \log (a-1)}{4 a}-\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(\frac{a+2}{a+1}\right)}{4 a}-\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(-\frac{a}{a+1}\right)}{4 a}-\frac{\pi \log (a-1) \log (a+2)}{4 a}-\frac{\pi \log (a+1) \log (a+2)}{4 a} & (1\le a\le3), \\ \color{Red}{-\frac{\pi \text{Li}_2\left(\frac{a^2}{a^2-1}\right)}{4 a}}-\frac{\pi \log \left(-\frac{a}{a+1}\right) \log \left(\frac{1}{1-a^2}\right)}{4 a}+\frac{\pi \text{Li}_2(-2)}{4 a}+\frac{\pi \text{Li}_2(2)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{1}{2}\right)}{2 a}+\frac{\pi \text{Li}_2\left(\frac{1}{1-a}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{1}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{a}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(-\frac{1}{a-1}\right)}{4 a}+\frac{\pi \text{Li}_2\left(\frac{1}{a+1}\right)}{2 a}+\frac{\pi \text{Li}_2\left(\frac{a}{a+1}\right)}{2 a}-\frac{\pi \text{Li}_2\left(-\frac{a}{2}\right)}{2 a}-\frac{\pi \text{Li}_2(1-a)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (1-a)}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a-2}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2\left(-\frac{3}{a-1}\right)}{4 a}-\frac{\pi \text{Li}_2(a-1)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a+2}{2 (a+1)}\right)}{4 a}-\frac{\pi \text{Li}_2\left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}+\frac{\pi ^3}{3 a}-\frac{\pi \log ^2(2)}{4 a}+\frac{\pi \log ^2(a+1)}{2 a}+\frac{i \pi ^2 \log (2)}{4 a}+\frac{\pi \log (2) \log (a)}{2 a}+\frac{\pi \log (2) \log (a+2)}{2 a}-\frac{\pi \log (2) \log (a-1)}{4 a}-\frac{\pi \log (2) \log (a+1)}{4 a}+\frac{i \pi ^2 \log \left(\frac{1}{1-a}\right)}{4 a}+\frac{\pi \log (3) \log (a-2)}{4 a}+\frac{\pi \log (a-2) \log (a-1)}{4 a}+\frac{\pi \log \left(\frac{a+2}{a+1}\right) \log \left(-\frac{1}{a+1}\right)}{4 a}+\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(-\frac{a}{a+1}\right)}{4 a}+\frac{\pi \log (a-1) \log (a+2)}{4 a}+\frac{\pi \log (a+1) \log (a+2)}{4 a}-\frac{\pi \log (a) \log (a+2)}{2 a}-\frac{\pi \log (a-2) \log \left(\frac{1}{a-1}\right)}{4 a}-\frac{\pi \log (3) \log \left(\frac{a+2}{a-1}\right)}{4 a}-\frac{\pi \log (2-a) \log (a-1)}{4 a}-\frac{\pi \log (a) \log \left(\frac{1}{a+1}\right)}{4 a}-\frac{\pi \log (3) \log \left(\frac{a-2}{a+1}\right)}{4 a}-\frac{\pi \log \left(\frac{1}{(a+1)^2}\right) \log \left(\frac{a+2}{a+1}\right)}{4 a}-\frac{\pi \log (3) \log (a+1)}{4 a}-\frac{\pi \log \left(-\frac{1}{a+1}\right) \log \left(\frac{a (a+2)}{(a+1)^2}\right)}{4 a}& (a\ge3). \end{cases}$$

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Update 3: Common fourier transforms $$\begin{aligned} &1.\int_{0}^{\infty}\operatorname{Ci}(x)\cos(\omega x)\text{d}x= \begin{cases} 0 &(0\le\omega<1), \\ \displaystyle{ -\frac{\pi}{4} }&(\omega=1), \\ \displaystyle{ -\frac{\pi}{2\omega} }&(\omega>1). \end{cases}\\ &2.\int_{0}^{\infty}\operatorname{Ci}(x)\sin(\omega x)\text{d}x= \begin{cases} \displaystyle{-\frac{\ln(1-\omega^2)}{2\omega}} &(0\le\omega<1), \\ \displaystyle{ +\infty }&(\omega=1), \\ \displaystyle{-\frac{\ln(\omega^2-1)}{2\omega} }&(\omega>1). \end{cases}\\ &3.\int_{0}^{\infty}\operatorname{Ci}(x)^2\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{ \frac{\pi\ln(1+\omega)}{2\omega} }&(0\le\omega\le2), \\ \displaystyle{ \frac{\pi\ln(\omega^2-1)}{2\omega} }&(\omega\ge2). \end{cases}\\ &4.\int_{0}^{\infty}\operatorname{si}(x)\sin(\omega x)\text{d}x= \begin{cases} 0 &(0\le\omega<1), \\ \displaystyle{ \frac{\pi}{4} }&(\omega=1), \\ \displaystyle{ \frac{\pi}{2\omega} }&(\omega>1). \end{cases}\\ &5.\int_{0}^{\infty}\operatorname{si}(x)\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{\frac{1}{2\omega}\ln\left ( \frac{1+\omega}{1-\omega} \right ) } &(0\le\omega<1), \\ \displaystyle{ +\infty }&(\omega=1), \\ \displaystyle{\frac{1}{2\omega}\ln\left ( \frac{\omega+1}{\omega-1} \right ) }&(\omega>1). \end{cases}\\ &6.\int_{0}^{\infty}\operatorname{si}(x)^2\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{ \frac{\pi\ln(1+\omega)}{2\omega} }&(0\le\omega\le2), \\ \displaystyle{ \frac{\pi}{2\omega}\ln\left ( \frac{\omega+1}{\omega-1} \right ) }&(\omega\ge2). \end{cases}\\ &7.\int_{0}^{\infty}\frac{\operatorname{Si}(x)}{x}\cos(\omega x)\text{d}x= \begin{cases} \displaystyle{-\frac{\pi}{2}\ln(\omega)} &(0<\omega\le1), \\ \displaystyle{0 }&(\omega\ge1). \end{cases}\\ \end{aligned}$$

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Definition: Functions $\operatorname{Si}_n(x)$ are defined by $$\operatorname{Si}_0(x)=\sin(x),\operatorname{Si}_n(x) =\int_{0}^{x} \frac{\operatorname{Si}_{n-1}(t)}{t}\text{d}t.$$ And we are able to get $$ \int_{0}^{\infty}\frac{\operatorname{Si}_2(x)\operatorname{si}(x)^2}{x} \text{d} x=\frac{7\pi^5}{1440}. $$

Answer 1

The following is an evaluation of $$\int_{0}^{\infty} \left( \operatorname{Si}(x)- \frac{\pi}{2} \right)^{3} \, \mathrm dx,$$ which was asked about in the comments and mentioned in Update.1.

Integrating by parts twice, we get $$\begin{align} \int_{0}^{\infty} \left(\operatorname{Si}(x) - \frac{\pi}{2} \right)^{3} \, \mathrm dx &= -\frac{3 \pi^{2}}{4} - 3\int_{0}^{\infty} \frac{\sin(2x)\left(\operatorname{Si}(x) - \frac{\pi}{2}\right)}{x} \, \mathrm dx \\ &= -\frac{3 \pi^{2}}{4} - 3 J, \end{align} $$

where

$ \begin{align} J&= -\int_{0}^{\infty} \frac{\sin (2x)}{x} \int_{1}^{\infty} \frac{\sin (xt)}{t} \, \mathrm dt \, \mathrm dx \\ & \overset{(1)}{=} -\int_{1}^{\infty} \frac{1}{t} \int_{0}^{\infty} \frac{\sin(2x)\sin (tx)}{x} \, \mathrm dx \, \mathrm dt \\ &= -\frac{1}{2}\int_{1}^{\infty} \frac{1}{t}\int_{0}^{\infty} \frac{\cos \left( (2-t)x\right) - \cos \left((2+t)x \right)}{x} \, \mathrm dx \, \mathrm dt \\ &\overset{(2)}{=} \frac{1}{2}\int_{1}^{\infty} \frac{1}{t} \ln \left(\frac{|2-t|}{2+t} \right)\, \mathrm dt \\ &= \frac{1}{2} \left(\int_{1}^{2} \frac{1}{t} \ln \left(\frac{2-t}{2+t} \right)\, \mathrm dt +\int_{2}^{\infty} \frac{1}{t} \ln \left(\frac{t-2}{2+t} \right)\, \mathrm dt \right) \\ & \overset{(3)}{=}\frac{1}{2}\int_{1}^{2} \left( \frac{1}{t} \ln \left(\frac{2-t}{2+t} \right)\, \mathrm dt + \int_{0}^{1/2} \frac{1}{v} \ln \left(\frac{1-2v}{1+2v} \right) \, \mathrm dv \right) \\ &= \frac{1}{2} \left( \int_{1}^{2}\frac{\ln(2-t)}{t} \, \mathrm dt- \int_{1}^{2} \frac{\ln(2+t)}{t} \, \mathrm dt + \int_{0}^{1/2} \frac{\ln(1-2v)}{v} \, \mathrm dv - \int_{0}^{1/2} \frac{\ln(1+2v)}{v} \, \mathrm dv \right) \\ & \overset{(4)}{=} \frac{1}{2} \left( \int_{1}^{2}\frac{\ln \left(1- \frac{t}{2}\right)}{t} \, \mathrm dt- \int_{1}^{2} \frac{\ln(1+ \frac{t}{2})}{t} \mathrm dt + \int_{0}^{1/2} \frac{\ln(1-2v)}{v} \, \mathrm dv - \int_{0}^{1/2} \frac{\ln(1+2v)}{v} \, \mathrm dv \right) \\ & \overset{(5)}{=} \frac{1}{2} \left( -\operatorname{Li}_{2} (1) + \operatorname{Li}_{2} \left(\frac{1}{2} \right) + \operatorname{Li}_{2} (-1) - \operatorname{Li}_{2} \left(- \frac{1}{2} \right)- \operatorname{Li}_{2} (1)+ \operatorname{Li}_{2} (-1) \right) \\ & \overset{(6)}{=} \frac{1}{2} \left(-\frac{\pi^{2}}{2} + \operatorname{Li}_{2} \left(\frac{1}{2} \right) - \operatorname{Li}_{2} \left(- \frac{1}{2} \right)\right) \\ & \overset{(7)}{=} \frac{1}{2} \left(-\frac{\pi^{2}}{2}+ 2 \operatorname{Li}_{2} \left(\frac{1}{2} \right)- \frac{1}{2} \operatorname{Li}_{2} \left(\frac{1}{4} \right) \right) \\ & \overset{(8)}{=} \frac{1}{2} \left(-\frac{\pi^{2}}{3}- \ln^{2}(2) - \frac{1}{2} \operatorname{Li}_{2} \left(\frac{1}{4} \right) \right). \end{align} $

Therefore, $$ \begin{align} \int_{0}^{\infty} \left(\operatorname{Si}(x)- \frac{\pi}{2} \right)^{3} \, \mathrm dx &= - \frac{3 \pi^{2}}{4} - \frac{3}{2} \left(-\frac{\pi^{2}}{3}- \ln^{2}(2) - \frac{1}{2} \operatorname{Li}_{2} \left(\frac{1}{4} \right) \right) \\ &= - \frac{\pi^{2}}{4} + \frac{3}{2} \ln^{2}(2) + \frac{3}{4} \operatorname{Li}_{2} \left(\frac{1}{4} \right). \end{align}$$

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$(1)$ Switching the order of integration is justified by Plancherel's theorem in the form $\int_{\mathbb{R}} f(x) \hat{g}(x) \, \mathrm dx = \int_{\mathbb{R}} \hat{f}(\omega) g(\omega) \, d \omega, $ which holds if $f$ and $g$ are square-integrable functions.

$(2)$ Generalized Frullani integral

$(3)$ Make the substitution $v= \frac{1}{t}$ in the second integral.

$(4)$ $\ln(2-t) = \ln(2) + \ln \left(1-\frac{t}{2} \right)$ and $\ln(2+t)= \ln(2)+ \ln \left(1+\frac{t}{2} \right)$

$(5)$ $-\int_{0}^{x} \frac{\ln(1-yt)}{t} = - \int_{0}^{xy} \frac{\ln(1-u)}{u} \, \mathrm du = \operatorname{Li}_{2}(xy)$

$(6)$ $\operatorname{Li}_{2}(1) = \zeta(2) = \frac{\pi^{2}}{6}$ and $\operatorname{Li}_{2}(-1) = - \eta(2) = - \frac{\pi^{2}}{12}$

$(7)$ Dilogarithm duplication formula

$(8)$ Special values of the dilogarithm

Answer 2

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(added on Jan.08, 22): Closed-forms of $\mathcal{L}(m,n)=\int_{0}^{\infty}\operatorname{Ci}(x)^m\operatorname{si}(x)^n\text{d}x$(notice: $\operatorname{si}(x)=\operatorname{Si}(x)-\frac\pi2$). \begin{align} &\left\{\begin{aligned} \mathcal{L}(1,0)=0,\qquad\mathcal{L}(0,1)=-1.\end{aligned}\right.\\ &\left\{\begin{aligned}\mathcal{L}(2,0)=\frac\pi2,\qquad \mathcal{L}(1,1)=\ln(2),\qquad\mathcal{L}(0,2)=\frac\pi2.\end{aligned}\right.\\ &\left\{\begin{aligned}&\mathcal{L}(3,0)=-\frac{3\pi}{2}\ln(2),\\ &\mathcal{L}(2,1)=-\frac34\operatorname{Li}_2\left ( \frac14 \right )-\frac32\ln(2)^2 -\frac{\pi^2}{12},\\ &\mathcal{L}(1,2)=-\frac{\pi}{2}\ln(2),\\ &\mathcal{L}(0,3)=\frac34\operatorname{Li}_2\left ( \frac14 \right )+\frac32\ln(2)^2 -\frac{\pi^2}{4}.\end{aligned}\right.\\ &\left\{\begin{aligned}&\mathcal{L}(4,0)=\frac{3\pi}{2}\operatorname{Li}_2 \left ( \frac{1}{4} \right )+3\pi\ln(2)^2+\frac{\pi^3}{4},\\ &\mathcal{L}(3,1)=\frac{15}{4}\operatorname{Li}_3\left ( \frac{1}{4} \right ) -\frac{5}{8}\zeta(3)+\frac{15}{4}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)+\frac{5}{2}\ln(2)^3 +\frac{\pi^2}{2}\ln(2),\\ &\mathcal{L}(2,2)=\frac{\pi^3}{12},\\ &\mathcal{L}(1,3)=-\frac{9}{4}\operatorname{Li}_3\left ( \frac{1}{4} \right ) +\frac{21}{8}\zeta(3)-\frac{9}{4}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)-\frac{3}{2}\ln(2)^3,\\ &\mathcal{L}(0,4)=-\frac{3\pi}{2}\operatorname{Li}_2\left ( \frac14 \right )-3\pi\ln(2)^2 + \frac{\pi^3}{4}.\end{aligned}\right.\\ &\left\{\begin{aligned}&\mathcal{L}(5,0)=-\frac{45\pi}{4}\operatorname{Li}_3\left ( \frac{1}{4} \right ) -\frac{15\pi}{16}\zeta(3)-\frac{45\pi}{4}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)-\frac{15\pi}{2}\ln(2)^3 -\frac{15\pi^3}{8}\ln(2),\\ &\mathcal{L}(3,2)=\frac{3\pi}{4}\operatorname{Li}_3\left ( \frac{1}{4} \right ) -\frac{29\pi}{16}\zeta(3)+\frac{3\pi}{4}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)+\frac{\pi}{2}\ln(2)^3 -\frac{\pi^3}{8}\ln(2),\\ &\mathcal{L}(1,4)=\frac{15\pi}{4}\operatorname{Li}_3\left ( \frac{1}{4} \right ) -\frac{55\pi}{16}\zeta(3)+\frac{15\pi}{4}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)+\frac{5\pi}{2}\ln(2)^3 +\frac{\pi^3}{8}\ln(2). \end{aligned}\right. \end{align}

Sketch of evaluation: The following two relations are true. $$\begin{aligned} &\mathcal{L}(3,1)-\mathcal{L}(1,3) = \frac{1}{4} \int_{0}^{\infty}\operatorname{Ei}(-x)^4\text{d}x=6\operatorname{Li}_3\left ( \frac{1}{4} \right ) -\frac{13}{4}\zeta(3)+6\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)+4\ln(2)^3+\frac{\pi^2}{2}\ln(2),\\ &\mathcal{L}(3,1)+\mathcal{L}(1,3) =3\int_{0}^{1}\frac{\ln\left ( 1+\frac{1}{x} \right ) }{(1+2x)^2} \left ( \frac{\pi^2}{12}-\ln(x)\ln(1+x)+\ln(x)^2 +\operatorname{Li}_2\left ( -\frac1x \right ) \right )\mathrm{d}x=\frac{3}{2}\operatorname{Li}_3\left ( \frac{1}{4} \right ) +2\zeta(3)+\frac{3}{2}\operatorname{Li}_2\left ( \frac{1}{4} \right )\ln(2)+\ln(2)^3+\frac{\pi^2}{2}\ln(2). \end{aligned}$$ The first equality is evaluated in here. The second one is mechanically evaluated by the MZIntegrate command.(In this question occured first)
Other relations: $$ \begin{aligned} &\mathcal{L}(5,0)+2\mathcal{L}(3,2)+\mathcal{L}(1,4)=-4\pi \left ( \mathcal{L}(3,1)+\mathcal{L}(1,3) \right ) ,\\ &\mathcal{L}(5,0)-2\mathcal{L}(3,2)-3\mathcal{L}(1,4) =-4\pi\left ( \mathcal{L}(3,1)-\mathcal{L}(1,3) \right ) . \end{aligned} $$ Since $\mathcal{L}(5,0),\mathcal{L}(1,3),\mathcal{L}(3,1)$ have closed-forms, $\mathcal{L}(3,2),\mathcal{L}(1,4)$ can be found.

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(add on Jan.10, 22): For $(4,1),(2,3),(0,5)$, we have no simple result. We need to compute some weight 4 logarithm integrals but we don't expect irreducible CMZVs. And as they do, $\mathcal{L}(m,n),m+n\ge6$ do not have good results either.

• If $m$ is even, then $(m,n)$ can be calculated by lower $(a,b),a+b=m+n-1$ with odd $b$. For example, to evaluate $(5,0),(3,2),(1,4)$, we use values of $(3,1),(1,3)$. This is generally true for higher $m+n$.
• To go further, we detect a sophisticated integral, $$ \small\int_{0}^{\infty}\left ( \operatorname{Ci}(x)^2+\operatorname{si}(x)^2\right )^2\left ( \operatorname{Ci}(x)^2+\operatorname{si}(x)^2+3\pi\operatorname{si}(x)\right ) \text{d}x =6\pi^3\operatorname{Li}_2\left ( \frac{1}{4} \right ) +12\pi^3\ln(2)^2. $$
• This only shows the simplest cases. Integrals $\int_{0}^{\infty} \operatorname{Ci}(x)\operatorname{Ci}(ax)\operatorname{Ci}(bx) \operatorname{Ci}(cx)\operatorname{Ci}(dx)\text{d}x$ no doubt have similar expressions. But it will be very long.